ph = pk a + log 10{[base]/[acid]}

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Bridget Newton
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FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.

1 FORMULA SHEET (tear off) N A = x C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) amu = x kg C = K K = C atm = 760 torr = 760 mm Hg 1 atm = bar pv = nrt R = L atm/mol K 1 L atm = J R = J/mol K 1 J= 1 kg m 2 /s 2 F = C/mol (1 volt) (1 Coulomb) = 1 Joule p A = X A p A [B] = k p B p A = X Bp A T b = K b m B T f = K f m B = [B]RT H = U + pv G = H - TS G rxn = G rxn + RT ln Q ln K = - G rxn/rt K p = K C (RT) n If ax 2 + bx + c = 0, then x = ( - b [b 2-4ac] 1/2 ) 2a K a K b = K w = 1.0 x (at T = 25 C) ph = pk a + log 10{[base]/[acid]} G = - nfe cell E cell = E cell - (RT/nF) ln Q ln K = nfe cell/rt

2 GENERAL CHEMISTRY 2 FOURTH HOUR EXAM Name Panthersoft ID Signature Part 1 (16 points) Part 2 (32 points) Part 3 (32 points) TOTAL (80 points) Do all of the following problems. Show your work.. 2

3 Part 1. Multiple choice. Circle the letter corresponding to the correct answer. There is one and only one correct answer per problem. [4 points each] 1) Calcium carbonate (CaCO 3) is a slightly soluble ionic compound. For which of the following solutions will the number of grams per liter of CaCO 3 that will dissolve be largest (at T = 25. C)? a) A M solution of hydrochloric acid (HCl) b) A M solution of sodium chloride (NaCl) A c) A M solution of sodium hydroxide (NaOH) d) A M solution of calcium nitrate (Ca(NO 3) 2) e) pure water 2) The half cell reduction potential for the process Cl 2(g) + 2 e - 2 Cl - (aq) is E = v at T = 25. C. The half cell oxidation potential for the process Cl - (aq) 1 / 2 Cl 2(g) + e - at T = 25. C, is a) v b) v B c) v d) v e) 0.00 v 3) Which of the following statements about galvanic cells is true? a) In a galvanic cell, oxidation occurs at the anode b) In a galvanic cell, oxidation occurs at the cathode D c) In a galvanic cell, the anode is negative and the cathode is positive d) Both a and c e) Both b and c 4) For a galvanic cell reaction to be spontaneous at standerd conditions which of the following must be true? a) G rxn = 0 and E cell = 0 b) G rxn < 0 and E cell < 0 C c) G rxn < 0 and E cell > 0 d) G rxn > 0 and E cell < 0 e) G rxn > 0 and E cell > 0 Part 2. Short answer. 1) Give the oxidation number for sulfur (S) for each of the following substances. [2 points each] SO 3 +6 H 2S -2 Fe 2(SO 4) 3 +6 S 8 0 2) What is the difference (if any) between a galvanic cell and an electrolytic cell? [5 points] In a galvanic cell a chemical reaction is used to generate a voltage. In an electrolytic cell, an external voltage is provided to force a chemical reaction to take place, usually in a direction it would not normally go. 3

4 3) Balance the following oxidation-reduction reactions for acid conditions. [10 points] H 2O 2(aq) + Cr 2O 7 2- (aq) O 2(g) + Cr 3+ (aq) ox H 2O 2(aq) O 2(g) + 2 H + (aq) + 2 e - x 3 red Cr 2O 2-7 (aq) + 6 e H + (aq) 2 Cr 3+ (aq) + 7 H 2O( ) net Cr 2O 2-7 (aq) + 3 H 2O 2(aq) + 8 H + (aq) 2 Cr 3+ (aq) + 3 O 2(g) + 7 H 2O( ) 4) Several standard reduction potentials are given below, at T = 25. C. Cu 2+ (aq) + 2 e - Cu(s) E = v Ni 2+ (aq) + 2 e - Ni(s) E = v Pb 2+ (aq) + 2 e - Pb(s) E = v Cd 2+ (aq) + 2 e - Cd(s) E = v For each of the following questions circle the correct answer. There is one and only one correct answer per problem. [3 points each] a) The substance that is the best oxidizing agent? Cd 2+ (aq) Cu 2+ (aq) Ni 2+ (aq) Pb 2+ (aq) b) The reaction with the largest value for the standard half-cell oxidation potential? Cd(s) Cd 2+ (aq) + 2e - Cu(s) Cu 2+ (aq) + 2e - Ni(s) Ni 2+ (aq) + 2e - Pb(s) Pb 2+ (aq) + 2e - c) The substance that will produce the most of grams of metal when electrolysis is carried out on the molten salt for 1.00 hour at i = 40.0 amp. CdCl 2 CuCl 2 NiCl 2 PbCl2 4

5 Part 3. Problems. 1) The solubility product for silver bromide (AgBr, MW = 187.8) is K sp = 7.7 x at T = 25. C. How many grams of AgBr will dissolve in L of pure water at T = 25. C? [10 points] The solubility reaction is AgBr(s) Ag + (aq) + Br - (aq) K sp = [Ag + ] [Br - ] = 7.7 x Initial Change Equilibrium Ag + 0 x x Br - 0 x x (x) (x) = 7.7 x x 2 = 7.7 x x = (7.7 x ) 1/2 = 8.8 x 10-7 M So the mass of AgBr that will dissolve in L of pure water is mass = 8.8 x 10-7 mol g = 1.65 x 10-4 g mol 5

6 2) Consider the following galvanic cell and half cell reduction potentials. Note the data and the problems both assume T = 25.0 C. Cu(s) Cu 2+ (aq) Fe 2+ (aq),fe 3+ (aq) Pt(s) Fe 3+ (aq) + e - Fe 2+ (aq) E = 0.77 v Cu 2+ (aq) + 2 e - Cu(s) E = 0.34 v a) What is the purpose of the Pt(s) in the above galvanic cell? [4 points] The Pt(s) serves as the site of the Fe 3+ /Fe 2+ reduction reaction. Pt is used as it will not itself be oxidized or reduced. b) Give the half cell oxidation reaction, the half cell reduction reaction, and the net cell reaction for the galvanic cell. [8 points] ox Cu(s) Cu 2+ (aq) + 2 e - E = v red Fe 3+ (aq) + e - Fe 2+ (aq) x 2 E = v net Cu(s) + 2 Fe 3+ (aq) Cu 2+ (aq) + 2 Fe 2+ (aq) E cell = v c) Find the value for E cell for the galvanic cell. [4 points] E cell = v (see above) d) The following ion concentrations are observed in the above galvanic cell [Fe 2+ ] = 2.1 x 10-4 M [Fe 3+ ] = M [Cu 2+ ] = 4.6 x 10-3 M What is E cell for the galvanic cell? [6 points] Q = [Cu 2+ ] [Fe 2+ ] 2 = (4.6 x 10-3 ) (2.1 x 10-4 ) 2 = 1.41 x 10-6 [Fe 3+ ] 2 (0.0120) 2 Fron Nernst, E cell = E cell - (RT/nF) ln Q = v (8.314 J/mol K)(298. K) ln(1.41 x 10-6 ) (2)( C/mol) = v ( v) = v 6

ph = pk a + log 10{[base]/[acid]}

ph = pk a + log 10{[base]/[acid]} FORMULA SHEET (tear off) N A = 6.022 x 10 23 C = ( 5 / 9) ( F - 32) F = ( 9 / 5)( C) + 32 1 amu = 1.661 x 10-27 kg C = K - 273.15 K = C + 273.15 1 atm = 760 torr = 760 mm Hg 1 atm = 1.013 bar pv = nrt